Solve 2cos⁡2x−3sin⁡x=0for0∘≤x≤360∘.

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bahaistag · Accepted Answer

So that we have only one type of trig function in this, let's use the fact that
$\cos^{2} (x) + \sin^{2} (x) = 1 \to \cos^{2} x = 1 - \sin^{2} x$
to convert that $\cos^{2}$ into an expression in $\sin^{2}$ .
$2 \cos^{2} x - 3 \sin x = 0 \Leftrightarrow 2 - 2 \sin^{2} x - 3 \sin x = 0 \Leftrightarrow 2 \sin^{2} x + 3 \sin x - 2 = 0$
Now here's the trick: let $y = \sin (x)$ . Then we have the quadratic equation
$2 y^{2} + 3 y - 2 = 0$
Using the quadratic formula, we can easily see that the two solutions for this are
$y = - 2, or, y = \frac{1}{2}$
Now substitute $y = \sin (x)$ back in to get
$\sin (x) = - 2, or, \sin (x) = \frac{1}{2}$
The first of those has no solutions. The sine function is always between -1 and 1. So we just need the solutions to the second equation. Looking at our unit circle, we see that they are
$x = 30^{\circ}, or, x = 150^{\circ}$

Solve 2cos2x−3sin x=0 for 0∘leq xleq360^∘.

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